Question

if 3x-4 is a factor of p (x) = 2x^3 - 11x^2 + kx - 20 ​

Answer 1

Step-by-step explanation:

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Answer 2

The value of k is -19.

If p(x) = and 3x-4 is a factor, then=$2x^3 - 11x^2 + kx - 20,$it means that p(x) can be written as:

p(x) = (3x-4)q(x)

where q(x) is a polynomial of degree 2 (since p(x) is of degree 3 and (3x-4) is of degree 1).

Expanding the right-hand side, we get:

p(x) = 3xq(x) - 4q(x)

We can compare this with the expression for p(x) given in the question to obtain:

$2x^3 - 11x^2 + kx - 20 = 3xq(x) - 4q(x)$

Simplifying this expression, we get:

$2x^3 - 11x^2 + kx - 20 = (3x - 4)q(x)$

Since q(x) is a quadratic polynomial, we can write it as:

q(x) = ax^2 + bx + c

where a, b, and c are constants.

Substituting this expression for q(x) into the previous equation, we get:

$2x^3 - 11x^2 + kx - 20 = (3x - 4)(ax^2 + bx + c)$

Expanding the right-hand side, we get:

$2x^3 - 11x^2 + kx - 20 = 3ax^3 + (3b - 4a)x^2 + (3c - 4b)x - 4c$

Equating the coefficients of the corresponding powers of x on both sides, we get:

$3a = 2 (coefficient of x^3)$

$3b - 4a = -11 (coefficient of x^2)$

$3c - 4b = k (coefficient of x)$

-4c = -20 (constant term)

Solving these equations simultaneously, we get:

a = 2/3

b = -5

c = -5

Therefore, q(x) = $2/3 x^2 - 5x - 5$, and the polynomial p(x) can be written as:

p(x) =$(3x-4)q(x) = (3x-4)(2/3 x^2 - 5x - 5)$

Expanding this expression, we get:

p(x) =$2x^3 - 11x^2 - 19x + 20$

Therefore, the value of k is -19.

For such more questions on factors,

https://brainly.in/question/6819403

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